Let $G$ be a quasisplit reductive group over a $p$-adic field $F$ with center $Z$, and let $\omega: Z \rightarrow \mathbb{C}^{\ast}$ be a quasicharacter. Let $f: G \rightarrow \mathbb{C}$ be a locally constant function satisfying $f(zg) = \omega(z)f(g)$ which is compactly supported modulo $Z$. This means that there exists a compact set $\Omega$ such that $\{ g \in G : f(g) \neq 0 \}$ is contained in the product set $Z.\Omega$.
Let $B$ be a Borel subgroup of $G$ with unipotent radical $U$. I'm trying to understand why for each $g \in G$ the integral
$$\int\limits_U f(ug)du$$
is absolutely convergent.
Some information which might be relevant: since $U$ consists of unipotent matrices and $Z$ consists of diagonalizable matrices, $U$ has trivial intersection with $Z$. Also, the product set $UZg$ is closed in $G$. This is because $UZ$ is a closed subgroup of $G$.
We have $f \in \textrm{c-Ind}_Z^G (\omega)$, so this answer shows that there exists $\phi \in C_c^{\infty}(G)$ such that
$$f(g) = \int\limits_Z \omega(z)^{-1}\phi(zg)dz$$
Then
$$\int\limits_U f(ug)du = \int\limits_U \int\limits_Z \omega(z)^{-1}\phi(zug)dz \space du$$
which is integration over the group $U \times Z = U.Z \subseteq G$. The support of the function $U.Z \rightarrow \mathbb{C}$, $x \mapsto \phi(xg)$ is $\textrm{supp}(\phi)g^{-1} \cap ZU$, which is compact, so the integral is absolutely convergent.